Ask question

Ask question

All categories

3 years ago

9

Which of the following ocean-floor features would contain the newest rocks?

2 answers:

Amiraneli [1.4K]3 years ago

7 0

The further away from the mid-ocean ridge, the older the rocks are on the ocean floor. Therefore, the ocean floor feature that would contain the newest rocks would be the mid-ocean ridge.

victus00 [196]3 years ago

4 0

Answer: Mid-ocean ridge

A mid-ocean ridge is an underwater mountain landmass generated after the plate tectonics. It forms when the ocean floor is uplifted by the magma generated by the rise of convection current during the divergent plate tectonics. The rocks generated by this process is due to the solidification of the lava after underwater volcanic eruption. Therefore, mid-ocean ridge likely to form new rocks in the oceanic bed.

You might be interested in

Answer the question based on this waveform.

Nuetrik [128]

Answer:

Cannot be determined from the given information

Explanation:

Given the following data;

Velocity = 24 m/s

Period = 3 seconds

To find the amplitude of the wave;

Mathematically, the amplitude of a wave is given by the formula;

x = Asin(ωt + ϕ)

Where;

x is displacement of the wave measured in meters.

A is the amplitude.

ω is the angular frequency measured in rad/s.

t is the time period measured in seconds.

ϕ is the phase angle.

Hence, the information provided in this exercise isn't sufficient to find the amplitude of the waveform.

However, the given parameters can be used to calculate the frequency and wavelength of the wave.

6 0

3 years ago

The picture below shows a wheelbarrow. Using a wheelbarrow can make it easier to lift a heavy object by

timofeeve [1]

Answer:

B IS CORRECT

Explanation:

8 0

3 years ago

Arturiano [62]

Explanation:

A light bulb changes electrical energy into <em>heat energy and light energy .</em>

4 0

3 years ago

Find the mass of a person who weighs 608 N

Anvisha [2.4K]

Answer:

$weight = mass \times accelaration \: due \: to \: gravity \\ 608 = x \times 10 \\ x = 608 \div 10 = 60.8 \\ mass = 60.8 \\ thank \: you$

7 0

3 years ago

Read 2 more answers

A solid conducting sphere with radius R carries a positive total charge Q. The sphere is surrounded by an insulating shell with

Illusion [34]

Answer:

Explanation:

Volume of the insulating shell is,

$V_{shell}=\frac{4}{3}\pi(R^3_2-R^3_1)$

Charge density of the shell is,

$\rho=\frac{Q_{shell}}{\frac{4}{3}\pi(R^3_2-R^3_1)}$

Here, $R_2 =2R, R_1 =R \,and\, Q_{shell} =-Q$

$\rho=\frac{Q_{shell}}{\frac{4}{3}\pi((2R)^3-R^3)}=\frac{-3Q}{28\piR^3}$

B)

The electric field is $E=\frac{1}{4\pi\epsilon_0}\frac{Qr}{R^3}$

For 0 <r<R the electric field is zero, because the electric field inside the conductor is zero.

C)

For R <r <2R According to gauss law

$E(4\pi r^2)=\frac{Q}{\epsilon_0}+\frac{4\pi\rho}{3\epsilon_0}(r^3-R^3)$

substitute $\rho=\frac{-3Q}{28\piR^3}$

$E=\frac{2}{7\pi\epsilon_0}\frac{Q}{r^2}-\frac{Qr}{28\piR^3}$

D)

The net charge enclosed for each r in this range is positive and the electric field is outward

E)

For r>2R

Charge enclosed is zero, so electric field is zero

8 0

3 years ago